Question

An electron, placed in an electric field, experiences a force \( F \) of \( 1 \text{ N} \). What are the magnitude and direction of the electric field \( E \) at the point where the electron is located (\( e = 1.6 \times 10^{-19} \text{ C} \))?

• A. $\frac{1}{e} \text{ N/C}$, $F$ and $E$ are along the same direction
• B. $\frac{1}{e} \text{ N/C}$, $F$ and $E$ are against each other
• C. $\frac{1}{e} \text{ N/C}$, $F$ and $E$ are perpendicular
• D. $e \text{ N/C}$, $F$ and $E$ are against each other
• E. $e \text{ N/C}$, $F$ and $E$ are perpendicular

Hint:

Always check the sign of the charge[ 14]. Positive charges experience force in the same direction as the field, while negative charges experience force in the opposite direction.

Answer 1

The Correct Option is B

Concept: In an electric field, the force $\vec{F}$ exerted on a charge $q$ is directly proportional to the electric field $\vec{E}$ at that location[ 14]. The relationship is defined by: $$\vec{F} = q\vec{E}$$ For a negative charge like an electron, the force vector acts in the direction opposite to the electric field vector.

Step 1: {Identify the known values and the charge of the particle.}
The magnitude of the force $F$ is given as $1 \text{ N}$[ 12]. The charge $q$ of an electron is $-e$, where $e = 1.6 \times 10^{-19} \text{ C}$.

Step 2: {Calculate the magnitude of the electric field $E$.}
Using the formula $E = \frac{F}{|q|}$: $$E = \frac{1}{|-e|} = \frac{1}{e} \text{ N/C}$$

Step 3: {Determine the directional relationship between $F$ and $E$.}
Because the electron is negatively charged ($q = -e$), the equation becomes $\vec{F} = -e\vec{E}$. The negative sign indicates that the force and the electric field are in opposite directions, or "against each other."